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A source of sound is moving with constant velocity of $20 \mathrm{~m} / \mathrm{s}$ emitting a note of frequency 1000 $\mathrm{Hz}$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (Speed of sound $v=340 \mathrm{~m} / \mathrm{s}$ )
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The correct answer is:
9:8
When source is approaching the observer, the frequency heard
$n_a=\left(\frac{v}{v-v_S}\right) \times n=\left(\frac{340}{340-20}\right) \times 1000=1063 \mathrm{~Hz}$
When source is receding, the frequency heard $n_r=\left(\frac{v}{v+v_S}\right) \times n=\frac{340}{340+20} \times 1000=944$ $\Rightarrow n_a: n_r=9: 8$
Short tricks: $\frac{n_a}{n_r}=\frac{v+v_S}{v-v_S}=\frac{340+20}{340-20}=\frac{9}{8}$.
$n_a=\left(\frac{v}{v-v_S}\right) \times n=\left(\frac{340}{340-20}\right) \times 1000=1063 \mathrm{~Hz}$
When source is receding, the frequency heard $n_r=\left(\frac{v}{v+v_S}\right) \times n=\frac{340}{340+20} \times 1000=944$ $\Rightarrow n_a: n_r=9: 8$
Short tricks: $\frac{n_a}{n_r}=\frac{v+v_S}{v-v_S}=\frac{340+20}{340-20}=\frac{9}{8}$.
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